SEBA Class 10 Maths Exercise 4.1 Solutions: Quadratic Equations | New Book 2026

Get free solutions to all Exercise 4.1 questions from SEBA Class 10 Maths. This Exercise 4.1 is from SEBA’s new book, 2026. We have solved all the questions in a simple way so that you can understand the concepts and get full marks in your upcoming Matric examination.

If you are a student of SEBA Class 10 who is using the SCERT textbook to study Maths, then you must have come across Chapter 4, Quadratic Equations. Here you can get complete SEBA Solutions for the New Book 2026 Class 10 Maths Chapter 4.1 Quadratic Equations in one place.

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SEBA Class 10 Maths Exercise 4.2 Solutions

Exercise 4.1

Q1: Check whether the following are quadratic equations:
(i) $(x+1)^{2}=2(x-3)$ 

Solⁿ :

The given equation is:

$(x+1)^{2}=2(x-3)$

$\Rightarrow x^{2}+2x+1=2x-6 $

$\Rightarrow x^{2}+1+6=0$

$\Rightarrow x^{2}+7=0$

It is of the form $ax^{2}+bx+c=0$, where $a \ne 0$ ($a=1, b=0, c=7$).

Therefore, the given equation is a quadratic equation.

(ii) $x^{2}-2x=(-2)(3-x)$

Solⁿ :

The given equation is:

$x^{2}-2x=(-2)(3-x)$

$\Rightarrow x^{2}-2x=-6+2x$

$\Rightarrow x^{2}-2x-2x+6=0$

$\Rightarrow x^{2}-4x+6=0$

It is of the form $ax^{2}+bx+c=0$.

Therefore, the given equation is a quadratic equation.

(iii) $(x-2)(x+1)=(x-1)(x+3)$

Solⁿ :

The given equation is:

$(x-2)(x+1)=(x-1)(x+3)$

$\Rightarrow x(x+1)-2(x+1)=x(x+3)-1(x+3)$

$\Rightarrow x^{2}+x-2x-2=x^{2}+3x-x-3$

$\Rightarrow x^{2}-x-2=x^{2}+2x-3$

$\Rightarrow -x-2x-2+3=0$

$\Rightarrow -3x+1=0$ or $3x-1=0$

It is not of the form $ax^{2}+bx+c=0$ ($a=0$).

Therefore, the given equation is not a quadratic equation.

(iv) $(x-3)(2x+1)=x(x+5)$

Solⁿ :

The given equation is:

$(x-3)(2x+1)=x(x+5)$

$\Rightarrow x(2x+1)-3(2x+1)=x^{2}+5x$

$\Rightarrow 2x^{2}+x-6x-3=x^{2}+5x$

$\Rightarrow 2x^{2}-5x-3=x^{2}+5x$

$\Rightarrow 2x^{2}-x^{2}-5x-5x-3=0$

$\Rightarrow x^{2}-10x-3=0$

It is of the form $ax^{2}+bx+c=0$.

Therefore, the given equation is a quadratic equation.

(v) $(2x-1)(x-3)=(x+5)(x-1)$

Solⁿ :

The given equation is:

$(2x-1)(x-3)=(x+5)(x-1)$

$\Rightarrow 2x(x-3)-1(x-3)=x(x-1)+5(x-1)$

$\Rightarrow 2x^{2}-6x-x+3=x^{2}-x+5x-5$

$\Rightarrow 2x^{2}-7x+3=x^{2}+4x-5$

$\Rightarrow 2x^{2}-x^{2}-7x-4x+3+5=0$

$\Rightarrow x^{2}-11x+8=0$

Since the above equation is of the form $ax^{2}+bx+c=0, a\ne0$ where $a=1$, $b=-11$ and $c=8$

Therefore, the given equation is a quadratic equation.

(vi) $x^{2}+3x+1=(x-2)^{2}$

Solⁿ :

The given equation is:

$x^{2}+3x+1=(x-2)^{2}$

$\Rightarrow x^{2}+3x+1=x^{2}-2 \cdot x \cdot 2+(2)^{2}$

$\Rightarrow x^{2}+3x+1=x^{2}-4x+4$

$\Rightarrow 3x+4x+1-4=0$

$\Rightarrow 7x-3=0$

It is not of the form $ax^{2}+bx+c=0$ ($a=0$).

Therefore, the given equation is not a quadratic equation.

(vii) $(x+2)^{3}=2x(x^{2}-1)$

Solⁿ :

The given equation is:

$(x+2)^{3}=2x(x^{2}-1)$

$\Rightarrow x^{3}+3 \cdot x^{2} \cdot 2+3 \cdot x \cdot (2)^{2}+(2)^{3}=2x^{3}-2x \quad [\because (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}]$

$\Rightarrow x^{3}+6x^{2}+12x+8=2x^{3}-2x$

$\Rightarrow x^{3}-2x^{3}+6x^{2}+12x+2x+8=0$

$\Rightarrow -x^{3}+6x^{2}+14x+8=0$

It is not of the form $ax^{2}+bx+c=0$ (it is a cubic equation).

Therefore, the given equation is not a quadratic equation.

(viii) $x^{3}-4x^{2}-x+1=(x-2)^{3}$

Solⁿ :

The given equation is:

$x^{3}-4x^{2}-x+1=(x-2)^{3}$

$\Rightarrow x^{3}-4x^{2}-x+1=x^{3}-3 \cdot x^{2} \cdot 2+3 \cdot x \cdot (2)^{2}-(2)^{3} \quad [\because (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3}]$

$\Rightarrow x^{3}-4x^{2}-x+1=x^{3}-6x^{2}+12x-8$

$\Rightarrow -4x^{2}+6x^{2}-x-12x+1+8=0$

$\Rightarrow 2x^{2}-13x+9=0$

It is of the form $ax^{2}+bx+c=0$.

Therefore, the given equation is a quadratic equation.

Q2: Represent the following situations in the form of quadratic equations:(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

Solⁿ :

Let the breadth of the rectangular plot be $x$ metres.

Then, its length should be $(2x+1)$ metres.

Now,

$\text{Area of the plot} = \text{length} \times \text{breadth} $
$ = (2x+1)x \text{ m}^{2} $
$= (2x^{2}+x) \text{ m}^{2}$

Given that area $= 528 \text{ m}^{2}$.

Therefore,

$2x^{2}+x=528$

$\Rightarrow 2x^{2}+x-528=0$

Therefore, the breadth of the plot satisfies the quadratic equation $2x^{2}+x-528=0$, which is the required representation of the problem mathematically.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Solⁿ :

Let the two consecutive positive integers be $x$ and $x+1$.

According to question,

$x(x+1)=306$

$\Rightarrow x^{2}+x=306$

$\Rightarrow x^{2}+x-306=0$

Therefore, the positive integer $x$ satisfies the quadratic equation $x^{2}+x-306=0$, which is the required representation of the problem mathematically.

(iii) Ram’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Ram’s present age.

Solⁿ :

Let Ram’s present age be $x$ years.

Then, his mother’s present age $= (x+26)$ years.

Ram’s age 3 years from now $= (x+3)$ years.

Mother’s age 3 years from now $= (x+26+3) \text{ years} = (x+29)$ years.

According to question,

$(x+3)(x+29)=360$

$\Rightarrow x^{2}+29x+3x+87=360$

$\Rightarrow x^{2}+32x+87-360=0$

$\Rightarrow x^{2}+32x-273=0$

Therefore, Ram’s present age satisfies the quadratic equation $x^{2}+32x-273=0$, which is the required representation of the problem mathematically.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solⁿ :

Let the uniform speed of the train be $x \text{ km/h}$.

Total distance to be covered $= 480 \text{ km}$.

Time taken to cover $480 \text{ km} = \frac{480}{x} \text{ hours}$.

If the speed is decreased by $8 \text{ km/h}$, then new speed $= (x-8) \text{ km/h}$.

Time taken with reduced speed $= \frac{480}{x-8} \text{ hours}$.

According to question,

$\frac{480}{x-8} = \frac{480}{x} + 3$

$\Rightarrow \frac{480}{x-8} – \frac{480}{x} = 3$

$\Rightarrow 480 \left( \frac{1}{x-8} – \frac{1}{x} \right) = 3$

$\Rightarrow 480 \left( \frac{x – (x-8)}{x(x-8)} \right) = 3$

$\Rightarrow 480 \left( \frac{8}{x^{2}-8x} \right) = 3$

$\Rightarrow \frac{3840}{x^{2}-8x} = 3$

$\Rightarrow 3(x^{2}-8x) = 3840$

$\Rightarrow x^{2}-8x = \frac{3840}{3}$

$\Rightarrow x^{2}-8x = 1280$

$\Rightarrow x^{2}-8x-1280=0$

Therefore, the speed of the train satisfies the quadratic equation $x^{2}-8x-1280=0$, which is the required representation of the problem mathematically.

Q3: For what value of p the equation $(p-2)x^{2}+3x+5=0$ can not be quadratic?

(a) 1

(b) 2

(c) -2

(d) 0 

Sol:

Given equation is $(p-2)x^{2}+3x+5=0$.

An equation of the form $ax^{2}+bx+c=0$ is not a quadratic equation when $a=0$.

Comparing with standard form, coefficient of $x^{2}$ is $(p-2)$.

For the equation to not be quadratic, we must have:

$p-2 = 0$

$\Rightarrow p = 2$

Hence, the correct option is (b).

Q4: Which of the following are quadratic equations?
(i) $(x+1)^{2}=2(x-4)$
(ii) $(x-3)(x+1)=(x+2)(x-3)$
(iii) $(x-2)^{2}+1=2x-4$
(iv) $x(x+3)+7=(x+2)(x-2)$ 

Choose the correct option:

(a) (i) and (iv)

(b) (i) and (ii)

(c) (i) and (iii)

(d) (ii) and (iv)

Solⁿ :

Checking each case:

For (i): $(x+1)^{2}=2(x-4)$

$\Rightarrow x^{2}+2x+1 = 2x-8$

$\Rightarrow x^{2}+9=0$ (It is a quadratic equation)

For (ii): $(x-3)(x+1)=(x+2)(x-3)$

$\Rightarrow x^{2}-2x-3 = x^{2}-x-6$

$\Rightarrow -x+3=0$ (It is not a quadratic equation)  F

or (iii): $(x-2)^{2}+1=2x-4$

$\Rightarrow x^{2}-4x+4+1 = 2x-4$

$\Rightarrow x^{2}-6x+9=0$ (It is a quadratic equation)

For (iv): $x(x+3)+7=(x+2)(x-2)$

$\Rightarrow x^{2}+3x+7 = x^{2}-4$

$\Rightarrow 3x+11=0$ (It is not a quadratic equation)

Thus, equations (i) and (iii) are quadratic equations.

Hence, the correct option is (c).